Question

complete the square and write the given equation in standard form. then give the center and radius of the circle and graph the equation.
$x^{2}+y^{2}+4x - 2y - 31 = 0$
the equation of the circle in standard form is $(x + 2)^{2}+(y - 1)^{2}=36$
(simplify your answer.)
the center of the circle is $(-2,1)$
(type an ordered pair.)
the radius of the circle is $r=$

Explanation:

Step1: Recall the standard form of a circle equation

The standard form of a circle equation is \((x - h)^2+(y - k)^2=r^2\), where \((h,k)\) is the center and \(r\) is the radius.

Step2: Complete the square for \(x\) and \(y\) terms

Given \(x^{2}+y^{2}+4x - 2y-31 = 0\).
Group \(x\) terms and \(y\) terms: \((x^{2}+4x)+(y^{2}-2y)=31\).
For \(x\) terms: \(x^{2}+4x=(x + 2)^2-4\) (using \((a + b)^2=a^{2}+2ab + b^{2}\), here \(a=x\), \(2b = 4\Rightarrow b = 2\)).
For \(y\) terms: \(y^{2}-2y=(y - 1)^2-1\) (using \((a - b)^2=a^{2}-2ab + b^{2}\), here \(a=y\), \(2b=2\Rightarrow b = 1\)).
Substitute back: \((x + 2)^2-4+(y - 1)^2-1=31\).
Simplify to get \((x + 2)^2+(y - 1)^2=36\).

Step3: Identify the center and radius

Comparing \((x + 2)^2+(y - 1)^2=36\) with \((x - h)^2+(y - k)^2=r^2\).
Center \((h,k)=(-2,1)\) (since \(x+2=x-(-2)\)).
Radius \(r=\sqrt{36}=6\) (because \(r^{2}=36\)).

Answer:

The radius of the circle is \(r = 6\)